Final answer:
The statement that the expression a^2 + a is even for any odd integer a is correct because when substituting the variable a with (2k + 1), where k is an integer, the final expression can be factored out to show a factor of 2, demonstrating its evenness.
Step-by-step explanation:
The question pertains to a statement regarding odd and even integers and their mathematical properties. Considering any odd integer a, the claim is that the expression a2 + a will always result in an even number.
We know that any odd integer can be written in the form of 2k + 1, where k is an integer.
Taking the square of an odd integer a, we get a2 = (2k + 1)2 = 4k2 + 4k + 1. Adding a (2k + 1) to a2 yields a2 + a = 4k2 + 4k + 1 + 2k + 1 = 4k2 + 6k + 2, which is clearly an even number since it can be expressed as 2(2k2 + 3k + 1).